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2x^2 + 11x + 128

Solve in factored form
a) (2x + 8)(x + 16)
b) (2x + 16)(x + 8)
c) (2x + 4)(x + 32)
d) (2x + 32)(x + 4)

1 Answer

5 votes

Final answer:

To solve the quadratic equation 2x^2 + 11x + 128 = 0 in factored form, we need to find two binomials that, when multiplied together, equal the original expression. None of the given factored forms are correct.

Step-by-step explanation:

To solve the quadratic equation 2x^2 + 11x + 128 = 0 in factored form, we need to find two binomials that, when multiplied together, equal the original expression. We can start by factoring out the common factor of 2:

2(x^2 + 5.5x + 64) = 0

Next, we need to find two binomials that can be multiplied together to give us x^2 + 5.5x + 64. Using the quadratic formula, we can find the roots of this quadratic equation:

x = (-5.5 ± √(5.5^2 - 4(1)(64)))/(2(1))

After simplifying the expression, we get:

x = (-5.5 ± √(30.25 - 256))/2

x = (-5.5 ± √(-225.75))/2

The square root of a negative number is not a real number, so this equation has no real roots. Therefore, none of the given factored forms (a), b), c), or d)) are correct.

User Litisqe Kumar
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